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Thursday, April 4, 2013

This theorem was called "Fundamental" at the time when the most important (if not the only) purpose of algebra was solving polynomial equations, and the higher degree - the better.
Nowadays algebra is concerned with much broader spectrum of problems, but the historical name of this theorem is retained.

The Fundamental Theorem of Algebra states that any polynomial equation has at least one solution in the field of complex numbers.

More precisely, assume we have an equation of the following type:
a[0]·z^n + a[1]·z^(n-1) + a[2]·z^(n-2) + ...
+ a[n-2]·z^2 + a[n-1]·z + a[n] = 0
where n is an integer, all coefficients ai are constant complex numbers, z is a complex variable the value of which we have to find and the coefficient a[0] at z^n is non-zero (so, the polynomial is truly of the nth degree).
Or, using Σ-notation,
Σa[i]·z^(n-i) = 0
where index i ∈ [0,n] (that is, an integer index i is changing in the interval from 0 to n).

This equation is called polynomial equation of nth degree in the field of complex numbers.

Then the Fundamental Theorem of Algebra states that such a polynomial equation must have at least one complex solution, oftentimes called the root of this polynomial.

Proof of this theorem is beyond the scope of this course (which is rather exception than a rule) because of its complexity. However, this theorem has a few interesting corollaries (simple consequences) that we will prove.

Let c be one of the roots of this polynomial, that is c is a solution of the corresponding polynomial equation, thus P(n)(c) = 0.

Then this polynomial of the nth degree can be represented as a product of (z - c) and some other polynomial of the (n-1)th degree:
P(n)(z) = (z-c)·P(n-1)(z)

Corollary 2
Any polynomial
P(n)(z) = Σa[i]·z^(n-i)
of the nth degree has exactly n roots in the field of complex numbers and can be represented as a product of expressions (z-c[i]) and a constant multiplier A, where i∈[1,n], c[i] are its roots and a multiplier A equals to the coefficient a[0] at z^n.

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